3
$\begingroup$

Prelude

This Post is a continuation of this Original Post. The original problem asked is:

How many solutions does the following equation have:

$$ a^x = \log_a(x) \,,\quad a \in (0,1) \wedge x \in\mathbb{R}^+_0 $$

And already has been answered (see Claude Leibovici's answer for details).

Observations

I found this problem interesting and I numerically investigated it a bit deeper. This led me to another question. Figures below show that such roots exist and are reals:

enter image description here enter image description here enter image description here

And roots can be computed numerically for several values of $a$:

enter image description here

My observations, so far, are:

  1. All roots must lie in $(0,1)$ because $\log_a(x) > 0 \, \forall x \in (0,1)$ and $\log_a(x) < 0$ elsewhere, and $a^x > 0 \,\forall x \in \mathbb{R}$;
  2. Solving this problem involves complex analysis (such as the use of Lambert W function) but the result will stay in the real domain;
  3. A "divergence" point occurs at $(e^{-e},e^{-1})$;
  4. Solving the original problem is equivalent to solve (base conversion and Lambert W properties):$$\ln(a) = \frac{W\left(x \ln(x)\right)}{x} = \frac{\ln(x)}{x} \Leftrightarrow a_k = \exp\left[\frac{W_k\left(x \ln(x)\right)}{x}\right]$$
  5. Roots become triple when $a < e^{-e}$ (dashed black vertical line, as shown by Claude Leibovici in his answer)
  6. Roots have asymptotic behavior, it can be checked in term of $a(x)$ for two branches:

    • one root tends to unity as $a\rightarrow 1$: $\lim\limits_{x\rightarrow 1} a = 1$ (green curve rightmost);
    • two roots tend to zero as $a\rightarrow 0^+$: $\lim\limits_{x\rightarrow 0^+} a = 0$ (green and orange curves leftmost).

Questions:

My main questions are:

  • How can I prove that one root tends to unity when base $a\rightarrow 0^+$ (blue curve leftmost)? By taking the limit of the proper branch.
  • Is the point 4 correct? Investigating the solution using Wolfram Alpha it seems both Leibovici and my expressions are equivalent. But the first form suffer a huge float arithmetic error with numpy library. Anyway they can be plotted using the latter form:

enter image description here

Side questions are:

  • How is called the point where branches diverge?
  • Do end of the branches also have a specific name?
  • Can we say that roots are multiple at the "divergence" point? If so, in what sense are they multiple? Claude Leibovici: Roots are multiple in the sense that three first degrees of Taylor expansion vanishes at $x=e^{-1}$ with $a=e^{-e}$.
  • Is the green branch a specific one because it behaves smoothly?
$\endgroup$
1
$\begingroup$

Probably not answering the questions but this is too long for comments.

Considering the function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ its derivatives are $$f^{(n)}(x)=a^x \log^n(a)+(-1)^n \frac{(n-1)!}{x^n\, \log(a)}$$ The first derivative cancels at two points given by $$x_1=\frac{W_0\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and}\qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ which, in the real domain, exist if $\frac{1}{\log (a)}\geq -\frac 1 e$ that is to say if $a \leq e^{-e}$. If this is the case, $f(x_1)<0$ and $f(x_2)>0$ which explains the three roots.

What is interesting is to look at what happens when $a = e^{-e}$. For this value, the solution of $f(x)=0$ is unique $x=\frac 1e$. At this point, the second derivative is also zero and the Taylor expansion is $$\frac{e^2}{6} \left(x-\frac{1}{e}\right)^3-\frac{5e^3}{24} \left(x-\frac{1}{e}\right)^4+O\left(\left(x-\frac{1}{e}\right)^5\right)$$ which makes that, at ths point, $x=\frac 1e$ is a triple root of the equation.

On another side, we could also solve the equation for $a$ and its solutions are given by $$a_1=\left(\frac{x \log (x)}{W_{0}(x \log (x))}\right)^{\frac{1}{x}}\qquad \text{and}\qquad a_2=\left(\frac{x \log (x)}{W_{-1}(x \log (x))}\right)^{\frac{1}{x}}$$ which do exist if $x \leq \frac 1e$. These two functions are worth to be plotted.

When $x \to 1$ the expansion of $a_1$ is $$a_1=1+(x-1)-(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right)$$ and using series reversion $$x= 1+(a_1-1)+O\left((a_1-1)^2\right)$$ making that if $x\to 0 \implies a_1 \to 0$.

$\endgroup$
  • $\begingroup$ Thank you for taking time to answer. I have updated my OP to integrate information you provided. I had problem to plot you expression of $a$ using numpy but I succeed with another form (looks like a float error propagation I think it is due to the fractional power that grows quickly as $x \rightarrow 0$). Anyway Wolfram seems to show that both our versions agree. $\endgroup$ – jlandercy Dec 4 '18 at 9:17

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.